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How does the form of (a+b)(a-b) relate to the binomial factor?

 Feb 4, 2015

Best Answer 

 #1
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+5

(a+b)(a-b)

this would solve out to $${{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{ab}}{\mathtt{\,\small\textbf+\,}}{\mathtt{ab}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{b}}$$ 

so... the $${\mathtt{ab}}$$(s) will cancel out and you will end with

$${{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{b}}$$ (but it'd be 2b not 2 + b)

 Feb 4, 2015
 #1
avatar
+5
Best Answer

(a+b)(a-b)

this would solve out to $${{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{ab}}{\mathtt{\,\small\textbf+\,}}{\mathtt{ab}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{b}}$$ 

so... the $${\mathtt{ab}}$$(s) will cancel out and you will end with

$${{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{b}}$$ (but it'd be 2b not 2 + b)

Guest Feb 4, 2015
 #2
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+3

The answer above is not correct. 

(a - b)(a + b)

$${{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{b}}}^{{\mathtt{2}}}$$        

 Feb 4, 2015

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